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Sequence of measurable subsets of [0,1] (Lebesgue measure, Measurable)

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\left\{E_{k}\right\}_{k\in N}[/itex] be a sequence of measurable subsets of [0,1] satisfying [itex]m\left(E_{k}\right)=1[/itex]. Then [itex]m\left(\bigcap^{\infty}_{k=1}E_{k}\right)=1[/itex].

    2. Relevant equations
    m denotes the Lebesgue measure.
    "Measurable" is short for Lebesgue-measurable.

    3. The attempt at a solution
    My intuition was to construct 2 disjoint and uncountable subsets of an uncountable set that all have measure equal to 1.

    After much struggling, I don't think this is possible. To be able to construct such a counterexample would mean there is a mapping of each subset to the natural numbers which would make them both countable.

    Are there non-constructive approaches to building this counterexample? My attempts haven't yielded anything meaningful. Each example (rational and irrational, transcendental and algebraic) I've considered ends up with measure 1 and measure 0 subsets instead of both measure 1.

    I am not sure if my counterexample building skills are lacking or I just need to adjust my intuition. Any help would be appreciated.
     
  2. jcsd
  3. May 29, 2013 #2

    Ray Vickson

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    What is the measure of the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##?
     
  4. May 29, 2013 #3
    If I can construct a counterexample of 2 disjoint and noncountable subsets on [0,1] of measure 1, then the measure of the complement of intersections would be 1. However, I have not been able to find such a counterexample. The answer based on what I have considered is 0.
     
  5. May 29, 2013 #4

    Ray Vickson

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    So, you do not intend to answer my question. Fair enough, but that's it for me.
     
  6. May 29, 2013 #5
    I answered your question the best I could.

    Your question is closely related to the question I am posing. If I could simply answer it, then I wouldn't've started the thread.
     
  7. May 29, 2013 #6

    D H

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    Try again, but this time answer the question. Start by answering this question: What is the complement (in [0,1]) of ##\cap_{k=1}^{\infty} E_k##? When you have answered this, can you find an upper bound on the measure of that complement?
     
  8. May 29, 2013 #7
    By DeMorgan, [itex]C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}[/itex]

    A lower bound immediately comes to mind. I'm still stewing on how to get an upper bound to pop out.

    Lower Bound:
    Consider: [itex]m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right][/itex]. [itex]m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]= m\left[0,1\right]-m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\geq m\left[0,1\right]-\Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=1-0=1[/itex]
     
  9. May 29, 2013 #8

    Office_Shredder

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    You have a set which is a subset of [0,1] and has measure bigger than or equal to 1, and you're not sure what its measure is?
     
  10. May 29, 2013 #9

    Ray Vickson

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    In Step 1 you did not follow through all the way: if ##E = \cap_k E_k,## then
    [tex] E^c = \bigcup_k E_k^c [/tex]
     
  11. May 30, 2013 #10
    Yep. I messed that up too. One more time.

    By DeMorgan, [itex]C\left(\cap_{k=1}^{\infty} E_k\right)=C\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]=\bigcup_{k=1}^{\infty}E^{c}_{k}[/itex]

    Consider: [itex]m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)[/itex]. [itex]m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\leq \Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=0[/itex]

    Therefore, [itex]m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)=0\Rightarrow m\left(\cap_{k=1}^{\infty}E_{k}\right)=1[/itex]
     
  12. May 30, 2013 #11

    Ray Vickson

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    Easier: from ##E = \cap_k E_k## we have ##E^c = \cup_k E_k^c## (no need for all those intermediate steps). Then [tex] 0 \leq m(E^c) \leq \sum_k m(E_k^c) = \sum_k 0 =0,[/tex] where the ##`\leq '## is from some standard property or result, depending on exactly what approach was used in the course/notes/textbook.
     
    Last edited: May 30, 2013
  13. May 30, 2013 #12
    I like the sandwhich approach better than mine.

    Thanks for the pointers. I really lost the forest from the trees on this problem.

    I got it stuck in my head that perhaps the irrationals were "dense enough" to create 2 subsets with the same measure. It didn't dawn on me to use L.measure properties to disprove the existence of that set.
     
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